Electronic Devices - Op-Amp Applications - Discussion

V1 is passed through an inverting opamp with resistors R3 and Rf.

So output voltage =-(Rf/R3)*V1.

Now the output of above stage is given as input to an inverting summer along with V2 as second input.

Here the output of the inverting stage is connected with R1 and
V2 is connected with resistor R2.

So final o/p=(- Rf/R2)V2 + (- Rf/R3)(- Rf/R1)V1.

Well, here the second term, [Rf/R3xRf/R1] contains two gains multiplied, so it is very obvious that two systems have to be cascaded [basic control system] and the first term can be easily provided by using the concept of superposition.

So, B seems to be right.

According to me 2 op-amp is required.

Anyone please explain how two op-amps are needed?

If output is equal to gain*gain then two op-amp is used.

Include two Rf => have two op-amp.

If we use a inverting configuration with input v1 and r1 and rf and pass the output of this op-amp to other op-amp which is also in the inverting configuration but is a summer. That means.

It has. Instead of r1 two resistances r2 and r3 inn parallel with input as v2 and other has output of the first op-amp as its input then we can get the required output.

Trick to do this--- it was a series of two output so we can make a summer circuit and since every term has rf/r1 so it ought to be a inverting amplifier. And the second term has multiplied term that means. The input voltage is (rf/r1) *v1 rather v1 which can be obtained from the previous amplifier.

I say, keenly observe the equation it contains two voltage sources. Therefore if we take two op amps connected serially then only we can multiply gains and achieve the equation as given in the question.

I would say .. only possibility of having 2 Rf (feedback resistance)terms is by having two feedbacks and that is only possible by having two op-amps with feedback.


correct me if I am wrong .. but I am pretty much sure..

HERE gain*gain. so 2 op-amp. are required.according to me..